The correct option is C 5
We have seen that ∣∣∫baf(x)dx∣∣≤∫ba|f(x)|dx But we are given that f(x)≥0∀xϵ(a,b)
So, | f(x)| = f(x) . And ∫baf(x)dx will be a positive value as f(x) is always positive in the given interval and also a < b (given). So basically both the modes are redundant.
.∫ba|f(x)|dx=∫baf(x)dx
&|∫baf(x)dx|=∫baf(x)dx
Thus, ∫ba|f(x)|dx=|∫baf(x)dx|=5
Note - If there was no information given about f(x) being positive in the interval and also about the limits
(a & b) then the answer could be 5 & 6 both. As |∫baf(x)dx|≤∫ba|f(x)|dx will always hold good.